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I've just found a very strange behavior. Let's consider following program:

{-# LANGUAGE NoMonomorphismRestriction #-}
{-# LANGUAGE TypeFamilies              #-}
{-# LANGUAGE ExtendedDefaultRules      #-}

module Main where

import Prelude
import Control.Monad.IO.Class


type family FuncArg m where
    FuncArg ((->) t) = 'Just t
    FuncArg m        = 'Nothing

test1 :: (MonadIO m) => m ()
test1 = do
    liftIO $ print 6

test2 :: (MonadIO m, FuncArg m ~ 'Nothing) => m ()
test2 = do
    liftIO $ print 6

main :: IO ()
main = return ()

The function tst1 compiles fine, while tst2 fails:

exe/Main.hs:21:14: error:
    • Could not deduce (Show a0) arising from a use of ‘print’
      from the context: (MonadIO m, FuncArg m ~ 'Nothing)
        bound by the type signature for:
                   tst2 :: (MonadIO m, FuncArg m ~ 'Nothing) => m ()
        at exe/Main.hs:19:1-49
      The type variable ‘a0’ is ambiguous
      These potential instances exist:
        instance Show Ordering -- Defined in ‘GHC.Show’
        instance Show Integer -- Defined in ‘GHC.Show’
        instance Show a => Show (Maybe a) -- Defined in ‘GHC.Show’
        ...plus 22 others
        ...plus six instances involving out-of-scope types
        (use -fprint-potential-instances to see them all)
    • In the second argument of ‘($)’, namely ‘print 6’
      In a stmt of a 'do' block: liftIO $ print 6
      In the expression: do { liftIO $ print 6 }

exe/Main.hs:21:20: error:
    • Could not deduce (Num a0) arising from the literal ‘6’
      from the context: (MonadIO m, FuncArg m ~ 'Nothing)
        bound by the type signature for:
                   tst2 :: (MonadIO m, FuncArg m ~ 'Nothing) => m ()
        at exe/Main.hs:19:1-49
      The type variable ‘a0’ is ambiguous
      These potential instances exist:
        instance Num Integer -- Defined in ‘GHC.Num’
        instance Num Double -- Defined in ‘GHC.Float’
        instance Num Float -- Defined in ‘GHC.Float’
        ...plus two others
        (use -fprint-potential-instances to see them all)
    • In the first argument of ‘print’, namely ‘6’
      In the second argument of ‘($)’, namely ‘print 6’
      In a stmt of a 'do' block: liftIO $ print 6

Giving explicit types to literals fixes the problem.